题目链接：

http://codeforces.com/problemset/problem/208/C

C. Police Station

time limit per test：2 secondsmemory limit per test：256 megabytes
#### 问题描述
> The Berland road network consists of n cities and of m bidirectional roads. The cities are numbered from 1 to n, where the main capital city has number n, and the culture capital — number 1. The road network is set up so that it is possible to reach any city from any other one by the roads. Moving on each road in any direction takes the same time.
>
> All residents of Berland are very lazy people, and so when they want to get from city v to city u, they always choose one of the shortest paths (no matter which one).
>
> The Berland government wants to make this country's road network safer. For that, it is going to put a police station in one city. The police station has a rather strange property: when a citizen of Berland is driving along the road with a police station at one end of it, the citizen drives more carefully, so all such roads are considered safe. The roads, both ends of which differ from the city with the police station, are dangerous.
>
> Now the government wonders where to put the police station so that the average number of safe roads for all the shortest paths from the cultural capital to the main capital would take the maximum value.
#### 输入
> The first input line contains two integers n and m (2 ≤ n ≤ 100, ) — the number of cities and the number of roads in Berland, correspondingly. Next m lines contain pairs of integers vi, ui (1 ≤ vi, ui ≤ n, vi ≠ ui) — the numbers of cities that are connected by the i-th road. The numbers on a line are separated by a space.
>
> It is guaranteed that each pair of cities is connected with no more than one road and that it is possible to get from any city to any other one along Berland roads.
#### 输出
> Print the maximum possible value of the average number of safe roads among all shortest paths from the culture capital to the main one. The answer will be considered valid if its absolute or relative inaccuracy does not exceed 10 - 6.
#### 样例
> **sample input**
> 4 4
> 1 2
> 2 4
> 1 3
> 3 4
>
> **sample output**
> 1.000000000000

题意

给你n个点（编号为1到n），m条边的有向图（无环，无重边，每个点都与其他点连通），现在要在一个点上建一个警察局，一条边，只要有一端与警察局相连，它就是安全的路，否则它就是不安全的路，现在问在哪个点建警察局能使从1到n的每条最短路上平均的安全路的条数（sigma(每条最短路上的安全路条数)/不同的最短路的条数）最多。

题解

首先跑一遍单源最短路求出所有最短路组成的DAG图g1，然后在DAG上dp两次，其中dp[i]表示从n到i的最短路条数，dp2[i]表示从1到i的最短路条数。

这样不同的最短路的条数就等于dp[1]或dp2[n]；

由于只有100个点，所以我们暴力枚举在2到n-1哪个点设警察局（在起点和终点设的话，平均的安全路的条数刚好会等于1)。每个警察局只会影响到它的邻边，我们现在把邻边分为两类（在g1中），一类以x为终点的边，一类为以x为起点的边，枚举所有的边，对于第一类边(u,x)，贡献值为dp2[u] * dp[x]。

对于第二类边(x,v)，贡献值为dp2[x] * dp[v]。sigma(每条最短路上的安全路条数)=sigma(每条边的贡献值）。

所以最后的答案就是max(sigma_i(每条边的贡献值))/dp[1]。

(sigma_i表示第i条边设警察局时的sigma(每条边的贡献值),sigma表示求和符号）

代码

#include<iostream>

#include<cstdio>

#include<cstring>

#include<vector>

#include<algorithm>

#include<map>

#include<queue>

using namespace std;

typedef long long LL;

const int maxn = 222;

int n, m;

vector<int> G[maxn];

//g1表示从1到n的所有最短路组成的DAG图。

vector<int> g1[maxn], g2[maxn];

int inq[maxn], d[maxn];

void spfa(int s) {

	memset(d, 0x3f, sizeof(d));

	queue<int> pq;

	inq[s] = 0; d[s] = 0; pq.push(s);

	while (!pq.empty()) {

		int u = pq.front(); pq.pop();

		inq[u] = 0;

		for (int i = 0; i<G[u].size(); i++) {

			int v = G[u][i];

			if (d[v]>d[u] + 1) {

				d[v] = d[u] + 1;

				if (!inq[v]) inq[v] = 1, pq.push(v);

				g2[v].clear();

				g2[v].push_back(u);

			}

			else if (d[v] == d[u] + 1) {

				g2[v].push_back(u);

			}

		}

	}

}

void get_g1() {

	for (int u = 1; u <= n; u++) {

		//		printf("%d: ",u);

		for (int i = 0; i<g2[u].size(); i++) {

			int v = g2[u][i];

			//			printf("%d ",v);

			g1[v].push_back(u);

		}

		//		puts("");

	}

}

LL dp[maxn];//dp[i]表示从n到i之间的最短路数

LL dfs(int u) {

	if (dp[u]) return dp[u];

	for (int i = 0; i<g1[u].size(); i++) {

		int v = g1[u][i];

		dp[u] += dfs(v);

	}

	return dp[u];

}

LL dp2[maxn]; //dp2[i]表示从1到i之间的最短路数

LL dfs2(int u) {

	if (dp2[u]) return dp2[u];

	for (int i = 0; i<g2[u].size(); i++) {

		int v = g2[u][i];

		dp2[u] += dfs2(v);

	}

	return dp2[u];

}

LL solve() {

	LL ret = dp[1];

	for (int i = 2; i<n; i++) {

		LL sum = 0;

		for (int j = 0; j<g1[i].size(); j++) {

			int v = g1[i][j];

			sum += dp[v] * dp2[i];

		}

		for (int j = 0; j<g2[i].size(); j++) {

			int v = g2[i][j];

			sum += dp2[v] * dp[i];

		}

		ret = max(ret, sum);

	}

	return ret;

}

int main() {

	scanf("%d%d", &n, &m);

	while (m--) {

		int u, v; scanf("%d%d", &u, &v);

		G[u].push_back(v);

		G[v].push_back(u);

	}

	spfa(1);

	get_g1();

	memset(dp, 0, sizeof(dp));

	dp[n] = 1;

	LL fenmu = dfs(1);

	memset(dp2, 0, sizeof(dp2));

	dp2[1] = 1;

	dfs2(n);

	LL fenzi = solve();

	printf("%.12lf\n", fenzi*1.0 / fenmu);

	return 0;

}

乱七八糟

统计安全边的时候各种奇葩的想法，组合数学没学好，硬伤orz。